In a chase scene, a movie stuntman runs horizontally off the flat roof of one building and lands on another roof 2.0 m lower. If the gap between the buildings is 4.8 m wide, how fast must he run to cross the gap?

where does the 4.9 come from

To determine the speed that the stuntman must run to cross the gap, we can use the principles of projectile motion.

Step 1: Determine the time of flight.
Given that the horizontal distance between the buildings is 4.8 m and the stuntman experiences vertical displacement of 2.0 m, we can determine the time it takes for the stuntman to cross the gap.
Use the formula for time of flight in projectile motion:

t = sqrt((2 * h) / g)

where h is the vertical displacement and g is the acceleration due to gravity (approximately 9.8 m/s^2).

Substituting the values:
t = sqrt((2 * 2.0) / 9.8)
t = sqrt(0.408)
t ≈ 0.638 s

Step 2: Calculate the required horizontal velocity.
Since the horizontal distance is given as 4.8 m, we can now determine the required horizontal velocity using the formula:

v = d / t

where v is the horizontal velocity and d is the horizontal distance.

Substituting the values:
v = 4.8 / 0.638
v ≈ 7.53 m/s

Therefore, the stuntman must run at a speed of approximately 7.53 m/s to cross the 4.8 m gap.

To find out how fast the stuntman must run to cross the gap between the buildings, we can use the principles of projectile motion and the equations of motion.

Step 1: Identify the known quantities:
- Initial vertical displacement (∆y) = -2.0 m (negative because he lands lower)
- Horizontal displacement (∆x) = 4.8 m
- Acceleration due to gravity (g) = 9.8 m/s^2

Step 2: Determine the time of flight:
Since there is no vertical acceleration (assuming air resistance is negligible), we can use the equation of motion for vertical displacement: ∆y = v₀y * t + (1/2) * (-g) * t²
Substituting the known values:
-2.0 m = 0 * t + (1/2) * (-9.8 m/s²) * t²
Simplifying the equation gives us: -4.9 t² = -2.0
Solving for t, we find: t = √(2.0 / 4.9) ≈ 0.642 s

Step 3: Calculate the horizontal velocity:
We can use the equation of motion for horizontal displacement: ∆x = v₀x * t
Rearranging the equation gives: v₀x = ∆x / t
Substituting the known values:
v₀x = 4.8 m / 0.642 s ≈ 7.49 m/s

Therefore, the stuntman must run at a speed of approximately 7.49 m/s to cross the gap between the buildings.

4.8 = u t

what is t, the time airborne?
2 = 4.9 t^2
so
t = sqrt(.408) = .639 second
so
u = 4.8/.639
= 7.51 m/s